How many timed does 2x^2-7x+5 touch or cross the x-axis

so you have the graph y = 2x^2-7x + 5. to touch the x axis, y= 0. Then:
2x^2 - 7x + 5 = 0

Since the graph is quadratic, it can have 2 roots so it touches the axis twice; 2 repeated roots, so it touches the x axis once; or imaginary roots, thus, the graph would not touch or cross the x axis.

To solve the problem:
2x^2 -7x + 5 = 0
(2x -5) (x - 1) = 0

then:
2x-5 = 0 and x-1 = 0

Thus: x = 2.5 and x = 1 are roots of the equation.
As the equation has two distinct roots, the curve y = 2x^2-7x + 5, touches the x-axis twice.

I assume you mean the graph of y = 2x^2 - 7x + 5 ? That touches/crosses rhe x-axis whenever y = 0. You can factorise that expression (or factor it if you're in the US/Canada) to (2x - 5)((x - 1) = 0 so y = 0 if x - 1 = 0 or if 2x - 5 = 0, so if x = 1 or 2x = 5 i.e. x = 5/2, so the graph crosses the x-axis twice, once at x = 1, then at x = 5/2. If the graph only just touched the x-axis, that would be a tangent to the curve, and we would have two equal roots (which I think in the US/Canada is called a root or solution of multiplicity 2, or words to that effect).

2x^2 - 7x + 5 = 0

x = (7 +/- √(49 - 4(2)(5))/4 = (7 +/- 3)/4

Since there are two x-intercepts, 2x^2 - 7x + 5 touches the x-axis twice.

The discriminant, and what it means, is all you need to answer this question.

D = b² - 4ac = 9

D > 0, therefore 2 points of intersection

Since the highest exponent is 2 that makes it a quadratic function. so it touches the x-axis at 2 different places.

twice